Với x = 0 , 0 ko là nghiệm của p/t ( L )
Với x khác 0 , ta có :
\(3x^4-4x^3-5x^2+4x+3=0\)
\(\Leftrightarrow3x^2-4x-5+\frac{4}{x}+\frac{3}{x^2}=0\) ( chia cả 2 vế cho x^2 )
\(\Leftrightarrow3\left(x^2+\frac{1}{x^2}-2\right)-4\left(x-\frac{1}{x}\right)+1=0\)
\(\Leftrightarrow3\left(x-\frac{1}{x}\right)^2-4\left(x-\frac{1}{x}\right)+1=0\)
Đặt \(x-\frac{1}{x}=a\) , ta có :
\(3a^2-4a+1=0\)
\(\Leftrightarrow\left(3a-1\right)\left(a-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=\frac{1}{3}\\a=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{x}=\frac{1}{3}\\x-\frac{1}{x}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{x^2-1}{x}=\frac{1}{3}\\\frac{x^2-1}{x}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x^2-3=x\\x^2-1=x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3\left(x^2-\frac{x}{3}-1\right)=0\\x^2-x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-\frac{x}{3}-1=0\\\left(x-\frac{1}{2}\right)^2-\frac{5}{4}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\frac{1}{6}\right)^2-\frac{37}{36}=0\\\left(x-\frac{1}{2}\right)^2=\frac{5}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\frac{1}{6}\right)^2=\frac{37}{36}\\\left(x-\frac{1}{2}\right)^2=\frac{5}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pm\sqrt{37}+1}{6}\\x=\frac{\pm\sqrt{5}+1}{2}\end{matrix}\right.\)
Vậy ...
\(3x^4-4x^3-5x^2+4x+3=0\\ \Leftrightarrow3x^4-3x^3-x^3-3x^2+x^2-3x^2+x+3x+3=0\\ \Leftrightarrow3x^2\left(x^2-x-1\right)-x\left(x^2-x-1\right)-3\left(x^2-x-1\right)=0\\ \Leftrightarrow\left(x^2-x-1\right)\left(3x^2-x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\\3x^2-x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1\pm\sqrt{5}}{2}\\x=\frac{1\pm\sqrt{37}}{2}\end{matrix}\right.\)
